Question

hyperbola passes through the point P(√2, √3) and has foci at (± 2, 0). Then the tangent to this hyperbola at P also passes through the point

$left parenthesis negative square root of 2 comma space minus square root of 3 right parenthesis$
$left parenthesis 3 square root of 2 space comma space 2 square root of 3 right parenthesis$
$left parenthesis 2 square root of 2 space comma 3 space square root of 3 right parenthesis$
$left parenthesis square root of 3 comma square root of 2 right parenthesis$

Solution

left parenthesis 2 square root of 2 space comma 3 space square root of 3 right parenthesis

Equation of hyperbola is $straight x squared over straight a squared minus straight y squared over straight b squared space equals space 1$
foci is (±2, 0) hence ae = 2, ⇒ a²e² = 4

b² = a²(e² – 1)
∴ a² + b² = 4
Hyperbola passes through √2,√3
$therefore space 2 over straight a squared space minus 3 over straight b squared space equals space 1 space... space left parenthesis 2 right parenthesis On space sol space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis straight a squared space equals 8 space left parenthesis is space rejected right parenthesis space and space straight a squared space equals space 1 space and space straight b squared space equals 3 therefore space straight x squared over 1 minus straight y squared over 3 space equals 1 Equation space of space tangent space is space fraction numerator square root of 2 straight x over denominator 1 end fraction minus fraction numerator square root of 3 straight y over denominator 3 end fraction space equals 1 Hence space left parenthesis 2 square root of 2 comma 3 space square root of 3 right parenthesis space satisfy space it.$

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